Untitled - Stacks are the Stanford

Exam No.
>
13.
(20 points)
Give an effective algorithm to determine for an arbitrary context-free
grammar, whether all words generated by it contain an equal number of
occurrences of letters
a
and b
31
.
Exam #
SPRING
1975
COMPUTER SCIENCE
COMPREHENSIVE EXAM
Given
10 May
1975
noon
- 12:00
- 4:30
Polya Hal 1 Room 11 1
9:00
1:30
TO:
THOSE WHO TAKE THIS EXAM
a)
The exam contains questions drawn from six areas of computer science.
The total possible score is 360 or 60 points in each area.
b)
During the lunch break you are urged to relax. You may not consult
any references or colleagues or write drafts of answers.
c)
Write your exam ff in the upper right corner of each page.
you have all 31 pages of the exam.
d)
Show your work as partial credit will be given.
syl labus
c)
Grading Strategy: To pass this exam, you should not leave any subject
area completely blank
as there will be a minimum competence requirement
in each area. After that, the exam will be graded on a curve so you
should try to maximize your total score.
f)
Sign the honor code statement.
.
Be sure
This exam is open
-
In recognition of and in the spirit of the Honor Code, I certify
that I have neither received nor given unpermitted aid on this exam.
Signed
GOOD luck:
..♦_
The results of the written part will be available Thursday, May 15, at
9:00 in the department office, Polya Hall 253. The programming problem may
be picked up at that time and will be due Monday, May 19, at 5:00 p.m.
f
1
Exam ff
SPRING 1975 COMPUTER SCIENCE
COMPREHENSIVE EXAM
Systems
1)
2)
3)
4)
5)
Deadlock
LISP
Paging Concepts
Compiler Quickies
Process Synchronization
Hardware
1)
2)
3)
4)
Combinational Logic
J-K Flip-Flop
Binary Counter
Architecture Concepts
Theory of Computation
1)
2)
3)
Diagonal ization
Finite Automata
Context-Free Language Quickies
Algorithms &
1)
2)
3)
ALGOL 60 Problem
Tree Traversal
Sparse Matrix
Numerical
l)
2)
3)
4)
Data Structures
Analysis
Banded Linear Systems
Arithmetic and Error Analysis
The Rank Problem
Newton's Method
Artificial Intelligence
1)
2)
3)
*.
"Understanding" Systems
A.I. Concepts
A. I *Quickies
.
SYSTEMS
(5 minutes)
Deadlock
1.
Exam ff
Three parallel processes share four tape units which can be reserved
or released only one at a time.
units.
Each process needs at most two tape
Can a deadlock occur in this system?
_
*
3
Exam #
SYSTEMS
1
(10 minutes)
and Systems Question
LISP
The bindings of variables in a LISP program act quite differently
from those in ALGOL. Describe briefly the basic difference in the
philosophy of the two approaches to determining binding.
2
How would a paging environment influence your implementation of the
CONS (structure building) function?
*
4
Exam ff
SYSTEMS
(20 minutes)
Paging Concepts
1.
A process refers to five pages, A,B,C,D,E, in the following order:
A-B-C-D-A-B-E-A-B-C-D-E
Assume the system uses a FIFO replacement algorithm.
a.
Find the
number of page faults starting with an empty store and 3 page
frames.
4
page frames.
Repeat part "a" assuming a LRU replacement algorithm.
b.
2.
Same question for
Using the hardware mechanisms of current machines that support paging,
describe a practical approximation to an LRU. page replacement algorithm
3.
Under what circumstances will page sharing improve the performance of a
paged multiprogramming system?
*
5
SYSTEMS
Exam
#
Compi ler Quickies
(5 minutes)
1.
Discuss three ways of reducing the storage requirements needed for
using precedence techniques.
(5 minutes)
2.
Give an example of why a compiler writer may want to combine two
different parsing techniques in one compiler.
(5 minutes)
3.
Can precedence functions be assigned to this matrix?
answer. )
s
s
b
-
.
H
s,
2
hl> -J
*
6
(Justify your
SYSTEMS
Exam
#
(5 minutes)
4.
What are the pros and cons of typed and typeless languages?
examples of each.
7
Give
SYSTEMS
Exam ff
(5 minutes)
Process Synchronization
Describe a primitive process synchronization mechanism.
8
Exam #
HARDWARE
(20 minutes)
Combinational Circuit Problem
The boolean Fibonacci function f is defined to be TRUE for the binary
inputs corresponding to
il,
2, 3, 5, 8, 1 3 5
and FALSE otherwise except
that we don't care what value it has for inputs |7 and
i)
9] "
Give the minimal sum of products expression for f
I
1
/
ii)
1
Give the minimal product of sums expression for f.
A multiplexor is a circuit that can select information from one of
several input channels and route that input to a single output lead. The
input terminal is selected by a binary encoded address supplied to the
circuit. A multiplexor can be used as a universal switching function by
connecting a fixed 1 or a fixed 0 to each of its inputs.
9
■»
Exam #
HARDWARE
i
ii)
Implement fon this 16 input multiplexor.
iv)
Can you implement f using only this single 8-input multiplexor and
inverters?
(if yes, do it)
UJP'oTS
1
ftWm*
/
"
«
10
HARDWARE
J-K Flip-Flop
(b)
st
Exam
#
(10 minutes)
A logic diagram for a J-K master-slave flip-flop can be drawn in
outputs that are
terms of two S-R latches as shown below. Draw the (^ and Q2
inputs
shown.
obtained from the J, X, and CX
11
HARDWARE
Exam
#
(15 minutes)
Bi nary Counter
Give the logic diagram of a lf-bit synchronous binary counter us mg
J X master-slave flip-flops.
(In a synchronous counter all of the output
bi ts must change simultaneously.)
f.
%
%■.
12
*
HARDWARE
Exam §
(15 minutes)
Architecture Concepts
Slow machines usually do things sequentially which faster machines can
usually do at least partially in parallel.
For each of the following terms ,
describe briefly what it is and how its use can increase a computer's
execution rate.
a.
multiprocessing
b.
memory interleaving
c.
wide buses
d.
associative memory
*
13
Exam #
THEORY OF COMPUTATION
(30 minutes)
Diagonal i zation Problem
Diagonal ization is a technique made famous when Cantor proved that
The technique assumes a list of items
the real numbers are not countable.
and establishes the existence of a new item which differs in some way from
each item in the list and is therefore not in the list.
a.
Use diagonal ization to show that there can be no effective enumeration
of the total recursive (defined for all inputs) functions.
by assuming that there is such an enumeration.)
1
«
14
a
(HINT:
start
THEORY OF COMPUTATION
b.
It is possible to give a listing [C., C
sensitive grammars.
,
Exam
C
...J of all
#
context
Show that there exists a recursive set which is
not context sensitive.
15
Exam #
THEORY OF COMPUTATION
(20 minutes)
Fini te Automata
Given the finite state machine M below whose inputs can be interpreted
as
- Letter
D - Digit
X - Del imi ter
L
Decimal Point
and where F..M is the
set
of final states
fU-KS 7 ]
i
«
(a)
Write a regular expression for the strings which end at state
For example, the regular expression
the set of strings over the set
T.* " D*x
«
16
£o,lj* foOOj Jo,l}*
|O,IJ having
6.
denotes
three consecutive O's.
Exam #
THEORY OF COMPUTATION
(b)
Construct a non-deterministic machine which recognizes the reversal
of the language recognized by M.
(c)
(Draw its state diagram.
>.
Construct a deterministic machine which recognizes the reversal of
the language recognized.
(d)
Relate this problem to lexical analysis.
_>
9
r
*
17
.
Exam 0
THEORY OF COMPUTATION
(10 minutes)
Context-free Language Quickies
Consider the language defined by the grammar
S
-*
a
S -> A b B
A -* b
A -» S B c
B
-» A a S
B -> a S A
with start symbol
S
and terminal alphabet
[a,b,cj
1.
Prove that the language contains no strings of length
2.
Find all strings in the language of length 9, having the letter "c"
as their middle (fifth) character.
I
V
18
«
100.
ALGORITHMS S DATA STRUCTURES
60
Algol
Exam §
(30 minutes)
Problem
The original ALGOL
60 report
shown on the following page.
concluded with the example procedure
RK
After fifteen years have gone by, we think
we know a little more about control structure and style.
Suppose ALGOL
60
has been extended to include the following new
syntax
(basic statement) ::= (loop statement)
(loop statement) ::= loop (statement list) while (Boolean expression):
(statement list) repeat
::= (statement) j (statement list); (statement)
(statement list)
and the semantics that, if
J
and T
are statement lists and B
is a
Boolean expression, the loop statement
loop J while B : T repeat
is equivalent to
J>
if B then
begin T;
loop J while B : T repeat
end
RK procedure, using the new loop statement, in order to
eliminate unnecessary go to statements which tend to obscure the control
Rewrite the
structure of the program in its original form.
Important note:
etc.
Large segments of the program have been marked
OL ,(3
,
J
so that you need not recopy them; simply give your answer in terms of
these abbreviations, as "procedure RK c.
"
etc.
You need not understand
precisely what goes on inside those chunks of code in order to solve this
problem.
Don't make any changes to the actual sequence of computations,
and don't try to do any tricky things like eliminating the Boolean
variable --"out" etc.
Simply remove as many of the
go to
statements and
meaningless labels as you can by taking advantage of the new loop construct
*
19
,
Exam
ff
OL
c
r
BB: RKlST(x,y,H,x2,y2); RKlST(x2,y2,H,x3,y3)
for k : =1 step 1 until n do
if comp(yl[k], y3[k], eta)>eps then go to CC;
/ comment: comp (a, b, c) is a
function designator, the value of which is the
( absolute value of the difference of the mantissas of a and b, after the exponents
\ of these quantities have been made equal to the largest of the exponents of the
20
ALGORITHMS £ DATA STRUCTURES
Exam #
(10 minutes)
Tree Traversal
Traversal of a binary
tree
in comprehens ive-qual order (CQ order)
is defined as follows.
procedure CQ(t) ; binary tree t;
if t
± A
then
begin if right(t) ¥= A then
begin CQpeft (ri ght (t) ) ) ; visit(t); CQ(left(t));
vlsl t (right (t)); CQ(right(right(t))) ;
end else
begin visit(t) ; CQ(left(t)) end
end.
left (t) , right (t) denote
respectively the left and right subtrees of a nonnull binary tree t
In what order are the nodes of the following binary tree visited,
when it is traversed in CQ order? (State the single-letter names of
Here
A denotes
a null binary tree; and
the nodes as they are encountered.)
.
.
<
i
Answer
«
21
ALGORITHMS & DATA STRUCTURES
data structure would you use to represent a sparse 100 x 100
matrix (1 word entries), if you are told that it contains
(a) At most 5
(b) At most 50
(c) At most 500
(d) At most 5000
non-zero entries
non-zero entries
non-zero entries
non-zero entries
The operations you will have to do on the matrix are
(i) fetch element A Ci,jl, given i and j
(ii) store element A L'..o> given i and j
(iii) fetch all non-zero elements of a given row
(iv) fetch all non-zero elements of a given column
...
#
(20 minutes)
Sparse Matrix
.What
Exam
22
Exam #
NUMERICAL ANALYSIS
Banded Linear Systems
1 )
(lO minutes)
It is well known that linear systems
Ax = b
can be solved using
Gaussian elimination without pivoting if the matrix
(e.g.,
A
is irreducibly diagonally dominant or
A
A
has certain properties
is positive definite).
Explain why this is an important practical consideration
when dealing with
banded matrices
m
A
=
,
(a. .)
is small compared with
n
a.
,
.
=
0
| 'l-j |>
the rank of the matrix.
i
*
if
23
m
,
say, where
Exam #
NUMERICAL ANALYSIS
(10 minutes)
Ari thmetic and Error Analysis
2)
a.
Do floating point and real arithmetic have the same arithmetic
properties (e.g., associativity,
+
and
x commute, etc.)?
Prove your point.
b.
What advantage does inverse error analysis have over forward error
analysis?
24
"a.
NUMERICAL ANALYSIS
Exam #
The Rank Problem
3
It is easy to ye
a)
What does this indicate about the condi tioning of the matrix on the
left hand side?
b)
Does the above example say anything about the use of Gaussian elimination
to determine the rank of a matrix?
(Recall that Gaussian elimination can
be viewed as transforming a matrix to upper triangular form.)
*
25
NUMERICAL ANALYSIS
Exam
#
(20 minutes')
k)
NewtohVs
Method
Newton's method is one of the best known iterative techniques for evaluating
The iterative step of the method is given by
the root of a function.
xn+l
a.
x
f(x )
n
IrW)
v
n
The graph below shows a function
using Newton's method has just evaluated F
the position for the estimate
f
near one of its roots.
at
x
as Indicated.
A program
Show geometrically
x
/
Can you predict any difficulty that will beset the program during the next few
iterations ?
»
«
26
'J
Exam #
NUMERICAL ANALYSIS
Some polynomials, especially those of high degree, are unstable
b.
in the sense that small changes in the coefficients will lead to large
changes in the zero.
"■'
=
1)
where
A
Ar
=
Show that the relative root change
-a
r
n
*"*
an
axn
p(x)
-
n l
n
+
a.x
is the polynomial
Oz.i<n-l
2)
p(r)
3)
"a
"
=
0
,
i.e.
r is a root
is the coefficient being slightly changed.
■j
/
L
X
"&
I
o* f>
A*-/
(r+tt
Newton's method is sometimes recommended because of its quadrat I c
c.
convergence.
Herman Newfellow has observed that the corrections
p(x.) / p (x.) were reduced by a factor of about 1/2 for each of the fi
10 iterations of his program.
What order of convergence does this mdi cate?
How would you explain this to Herman?
i
27
*
rst
r
Exam #
ARTIFICIAL INTELLIGENCE
(30 minutes)
"Understanding Systems"
In recent years the A.I. research community has been describing much of
its research as "understanding":
viz. speech understanding, language
understanding, and image understanding.
l)
(20 minutes)
What A.I. concepts, capabilities, program organizations,
representations, and tools are used in "understanding" systems? Which
of these are relatively domain specific and which are general? Illustrate
and refine your answer by dealing specifically with one of the following
areas of program understanding.
For the area chosen, concentrate on
one system in detail or survey the few systems which exist in the area.
Keep your answers brief and to the point.
a)
English language understanding
b)
speech understanding
c)
organic-chemical mass spectral data interpretation
«
28
ARTIFICIAL INTELLIGENCE
2)
Exam ff
(10 minutes)
Suppose we designed an image understanding computer
system (ala A.I. vision systems) to ride along on a future ERTS
(Earth Resources Technology Satellite) orbiting mission and to radio
back to scientists on earth what it "understood" about the scenes
being viewed.
From the A.I. scientist's point of view, what would be
the essential design problem in making the system work?
T
■'-.
29
ARTIFICIAL INTELLIGENCE
A. I
.
Exam #
(20 minutes)
Concepts
Define each of the following problem-solving concepts and methods, and
give a short (one or two sentence) description of the sort of problem
(For instance, "table lookup" might be
defined as the selection of data elements from an Information structure by
processing keys associated with the data. It would be applicable in
situations where the set of possible answers is explicit, there are
appropriate keys and selection functions, and the number of elements is
small enough to allow the storage and retrieval to be reasonably efficient )
domain to which it is applicable.
1
.
heuristic search
2.
means-ends analysis
3.
hill cl imbing
k.
"demons"
5.
production systems
*
30
Exam ff
ARTIFICIAL INTELLIGENCE
A.I
1
(10 minutes total)
Quickies
The state-space approach to problem solving is a special case of the
problem reduction approach.
True or False?
why bother dealing with a separate concept
2
(True or False).
(if true, then
of "state-space methods"?)
Explain.
The alpha-beta technique is often faster than the
minimax technique but it may achieve a less optimal answer.
3
It has been suggested that work on programs of a general problem solving
nature has been shelved temporarily.
would be the reason for this trend?
31
#
Supposing this is correct, what
State your answer briefly.
*
15 MAY 1975
i
COMPUTER SCIENCE COMPREHENSIVE
PROGRAMMING PROBLEM
Simulation of a Packet-switching Network
(Due Tuesday, May 20, at 9.00 a.m. in CS Dept. Office)
Computer networks are becoming more widespread and are influencing the way people can .use
computers. Although networks seem conceptually simple, some aspects of their design were not
appreciated until the networks were actually implemented. In the problem you will be able to
experiment with some elements of network design as well as learn something about the programming of
a simulation.
Background
The basic model for a network that we have in mind is a simplified version of the ARPA network.
The network is made up of IMPs, which are minicomputers used for message switching and which
are connected to each other by transmission lines. Each IMP is also connected to a HOST which is
a local computer where the user does his actual computations. Each IMP is in fact a computer and is
busy managing the message traffic between HOSTs. For the purposes of this simulation, we will be
mostly concerned about whrjt is going on inside the IMPs and will use a relatively simple model for
the activity of each HOST. The traffic on the network consists of variable length messages. A
message enters the network at the source (or sending) HOST, which specifies a destination (or
receiving) HOST. The message is transmitted through an appropriate set of IMPs and finally leaves
the network at a destination HOST.
-
MESSAGES
Messages have the following attributes:
1) source: HOST or IMP which sent it
2) destination: HOST or IMP to receive it
?.) acknowledge bit: Bit which is set if the message is simply an
acknowledgement of a previous message
4) length: number of bits in message.
Specified by sender.
- used
5) sequence number: set by sender (32 or 36 bits)
messages are delivered in the order they are sent.
-
to assure that
For simplicity, we will assume that messages are sent as a whole we will not model the packing and
unpacking of messages into smaller units. Also, we will pay no attention to the contents of messages.
For our purposes we will assume that there are only two sorts of messages:
1) Acknowledgement messages, which have the acknowledge bit set
and have length HEADER BITS.
*
I
2) Data messages, which come in two flavors:
a) short (length = HEADERBITS+SHORTBITS)
b) long (length = HEADERBITS+LONCBITS)
Short messages occur with probability SHORTPROB and long messages
occur with probability ( 1 SHORTPROB).
-
LINES
Lines carry the messages between IMPs or between an IMP and a HOST. The lines between IMPs
carry messages at a fixed rate, LINERATE, in thousands of bits per second. They are full-duplex,
that is, they can carry messages simultaneously in both directions (at LINERATE in each direction).
The transmission time of a message is assumed to depend only on the message length. Furthermore,
each line connecting an IMP to another IMP is assumed to have a probability,
LOSSPROB.<Ieneth of messago of garbling a message so that it will have to be retransmitted.
(Yes, we know this cannot be quite right, but accept it as an approximation). Error detection
circuitry in the receiving IMP will detect broken messages.
The lines connecting an IMP to its HOST are assumed to have zero error and operate at
HOSTRATE: In addition, they have a special BLOCK signal by which the IMP can prevent the
HOST from sending more messages.
HOSTS
For our purposes, assume that each IMP is associated with exactly one HOST.
Each HOST generates messages with a POISSON' arrival time distribution. This means that the
intervals between the times that the HOST queues messages for transmission to its IMP are
exponentially distributed. These may be generated for the simulation by the relation
T
,=T -TLAMBDA.(In(r))
n+ 1 n
To =
°
where r is uniformly distributed on (0,1), and TLAMBDA = mean interanival time.
No HOST sends a message to itself. The distribution of each message from a HOST is equal to the
destination of the previous message sent from that HOST with probability SAMEPROB; otherwise,
the destination is randomly chosen from among the HOSTs other then the sending HOST
(including the HOST of the previous message), each with equal probability.
As messages are generated, the HOST places them into an output queue to its corresponding IMP.
The IMP accepts messages from the HOST at rate HOSTRATE unless the IMP is blocked (in
which case, the messages just remain in the HOST's output queue). If the HOST runs out of buffer
space, it stops generating messages until a buffer becomes free.
If an IMP blocks a HOST while a transfer (into the IMP) is in progress, that transfer will be
completed, but no new HOST-to-IMP transfers will be started until the IMP unblocks the HOST.
We assume that the HOST is always able to accept a message from its IMP at the HOSTRATE.
We will not simulate what the HOST does with the messages it receives.
*
I
.
IMPS
The IMP is the basic message switching processor of the system. Each IMP is connected to some
subset of the set of other IMPs as determined by the topology of the network.
When an IMP accepts a message from its corresponding HOST, it adds a sequence number telling
the order (first, second, third, etc) of that message in the stream from the source to the destination
HOST. (In other words, there is a separate set of sequence numbers for each HOST-HOST pair).
Internally, each IMP has a Route Control Table by which it determines how to route a message.
This is simply a table of line numbers indexed by destination IMP or HOST.
Every IMP maintains a buffer pool from which it allocates buffers for its various processes. For
simplicity, each buffer is assumed to be long enough to contain one maximum length message.
to keep a buffer available on each of its lines ready to receive input. When an
ungarbled message arrives, and there is a free buffer available, the IMP processes it by doing the
An IMP tries
following:
1. Send an acknowledgement. (This does not necessarily require a new
buffer to create the acknowledgement message).
2. If the message is for its HOST, put the message on the output queue for
transmission to the HOST. Otherwise, look-up the destination line in the
Route Control Table and put the message on the output queue for that
line.
If there is no buffer available when a message arrives, the message is lost.
After a message has been sent off (to another IMP), the IMP cannot reclaim the message buffer
until an acknowledgement has been received from the IMP at the other end of the line. If no
acknowledgment has been received for TIMEOUT seconds, the IMP should retransmit the message.
The
IMPs' processing
time to allocate buffers, etc., is assumed to be zero.
NOTE: Since some messages get delayed in the network due to line loss and re-transmission delays,
messages may arrive for an IMP's HOST out of sequence order. An IMP must deliver messages to
its HOST in correct sequence order. (This implies that the destination HOST never receives the
same message twice.)
.
THE PROBLEM
specified here the order and the timing by which the IMP sends messages. Part of
your design problem is to work out a scheme that avoids excessive delay. Be sure to explain your
We have
not
scheme in the documentation.
You are to build a simulator & simulate the performance of a sample network (see below) from time
zero until a simulated time DEADLINE. At time zero, all buffers and queues are empty. Each
HOST'S first message will be to a destination and will occur at time T. for that HOST.
1. Assume that HOSTs and IMPs have an infinite buffer pool.
*
I
Print
out
a distribution (histogram) and averages for each of the following quantities:
1. average number of occupied buffers per IMP per unit time. (Lump
together the. statistics for all IMPs, since each will have the same
distribution in the network considered)
2. average number of occupied HOST buffers per
unit time. (Again,
lump
together the statistics for each HOST.)
3. delay time for messages between each class of HOST-HOST pairs. (In
the sample data set, there are just three such classes.) Lump together all
statistics within a given class. You should calculate delay time from when
the message is created in the source HOST to when transfer into the
destination HOST is complete.
.
Print out these statistics at every REPORTINGJNTERVAL of simulated time. Clear out the
accumulated totals so that each interval has fresh statistics.
2. Same question assuming that each IMP has only IMPSPACE buffers available and every HOST
has only HOSTSPACE buffers available.
3. Document your program. In
items.
particular, describe and explain your implementation of the following
a. representation of HOSTs and IMPs
b. handling of time in the simulation
c. data structures used in the program
d. IMP buffer allocation schemes
4. Have you thought about what you have done? Explain how you would redesign these network
and your implementation of the simulation to overcome peculiarities and difficulties you
observed in this problem. Did the test data set fully expose the capabilities and pathologies of your
algorithms
network algorithms?
Programs will be graded according to the following criteria:
a. Functionality. Are the results correct?
b. Clarity. Is the program and documentation clear? Is the program
easy to modify?
c. Appropriateness of data structures.
t
Partial credit will be given for incomplete solutions, but be sure
done, and indicate how you would do the rest.
«
to
document fully what you have
»
TEST DATA: Run your program for the following data
HOSTRATE: =IBB kb
SHORTBITS := 128
LINERATE:=SB kb
HtAUHHBITS := 9G
SHORTPROB := .35
TI MEOUT: = .05 sec
OEADLINE:=2 sec
L0SSPR0B:=.8881 per bit
LONGBITS := 1088
SAHEPROB := 8.75
HOSTSPACE : =I
=I B
TLAHBDA: -.886 sec
ItIPSPACE:- 25
HIPSPACE:=
REPORTING_INTERVAL=B.S sec
LIST OF CONNECTIONS
2
3
5
3
G
4
5
G
G
1
1
1
2
2
3
4
4
5
Route control
table
Ron
Classes of
A = I 1«2,
B = I 1-5,
C = I 1»4.
4
pairs
1
2
3
4
5
G
x
2
3
3
B
2
1
x
3
3
1
G
1
2
x
4
1
2
5
G
3
x
5
G
1
6
4
4
x
G
S
2
4
4
5
x
2«3, 3-1, 4«5, 5«6, 6«4 I
2«6, 3«4 1
1«G, 2»4, 2*5, 3*5, 3»G 1
ANSWERS
SPRING 1975 COMPREHENSIVE EXAM
SYSTEMS
1.
Deadlock
No
- A deadlock
can only occur if all the tape units are reserved and
processes are waiting indefinitely for resources which will never be
If all four tape units are reserved, then one of the
available.
processes has all the tape units it needs and Its two units will become
available when it finishes. (Partial credit was given to persons who
interpreted the problem as allowing deadlock to specific tapes (not
tape units) or other resources if their answer indicated clearly that
they understood deadlock.)
2.
LISP
1.
In ALGOL, variable bindings are static and semantical ly determined
at compile time from the block structure and scope rules.
In LISP,
bindings are dynamic and depend on the order that routines are
called.
A LISP variable is bound to its most recent definition on
a value stack.
2.
Try to place the new cell on the same page as the old structure
CAR).
(either the CDR or the
This is to alleviate paging during
traversal of the data structures and implies that a free storage
list should be associated with each page.
3.
Paging Concepts
1
.
a.
(* indicates a page fault)
FIFO with 3 frames.
A-B-C-D-A-B-E-A-B-C-D-E
JL
JL
JL
JL
J-
9 faults
J-
FI FO with 4 frames
A-B-C-D-A-B-E-A-B-C-D-E
JL
b.
JL
JL
JL
LRU with
JL
JL
JL
JL
JL
JL
JL
JL
JL
JL
JL
J*;.
JL
-L
ABCDABEABCDE
ABCDABEABCD
ABCDABEABC
*
(FIFO Anomaly)
frames (List shown)
3
A-B-C-D-A-B-E-A-B-C-D-E
JL
10 faults
10 faults
Page 2
LRU with
4
frames
8 faults
A-B-C-D-A-B-E-A-B-C-D-E
JL.
JL
JL
J-
JL
JL
JL
JL
JL
JL
JL
ABCDABEABCDE
ABCDABEABCD
ABCDABEABC
ABCDDDEAB
2.
Associate a reference bit with each page which is set with each
reference and which Is cleared later by a pointer advance.
When it
to select a page for removal, a pointer advances in a
is time
circular list of pages and selects the first page whose reference
bit is clear. (while clearing and skipping those set).
Another scheme (which is only available on a single machine) is to
associate an aging register with each page..
The register is set to
a maximum value at each reference and is decremented by an interval
timer.
removal
3.
.
At a fault, the page with the lowest counter is selected for
-
e.g.
On a system where there are many commonly used programs
compilers and editors
it is useful to reduce paging by having users
-
share the common pages.
Such sharing reduces both the overhead of
moving code to and from the drums and the required amount of drum
Sharing data pages between programs can be much more
efficient than other methods of transferring data from producers to
storage.
consumers.
4
Compi ler Quickies
1.
The idea is to reduce the storage required for the precedence matrix.
storing only the
Three methods are (l) sparse matrix techniques
-
non-blank relations (2) precedence functions (if they exist) (3) pointers
to identical rows or columns.
2.
Most compilers use a simpler technique for the scanner than for the
context-free parser.
Some compilers use top down methods for good
ri
error recovery but switch to operator precedence on expressions to
avoid backup. In FORTRAN, a special parser might be used for FORMAT
statements.
3.
No. From the matrix we conclude that
g(s-)
=
f(s.)>g(s ) = f(s )>g(s )
2
1
2
which is unsat isf iable in precedence functions.
*
4.
Typed languages
-
Page 3
e.g. ALGOL, SAIL, FORTRAN, PASCAL, SIMULA
Advantages
1) Many more errors detectable at compile time
2) Compiler knows more about what a program is doing and can
perform optimizations.
3) System provides
type
conversion automatically.
Disadvantages
1) Less flexible language.
2) Compiler harder
Typeless languages
to
- e.g.
write
BCPL, LISP, APL, EULER, GEDANKEN
Advantages
l) More flexibility possible in programming resulting sometimes
in
greater efficiency.
2) Compiler easier due
to the uniform treatment of addresses
Disadvantages
1) Harder to debug.
Remarks:
Some bugs result in subtle smashing of code or data
LISP has untyped variables, but typed values.
In languages
like PASCAL where there are user defined data types, it is
possible to do more representation independent programming.
4
Process Synchronization
The Test-and-Set instruction provides a primitive way for a process to
check a flag before proceeding.
A process l) waits for a flag to be
clear, 2) sets the flag, 3) enters its critical region and clears the
flag
when it is done.
The Test-and-Set instruction guarantees mutual
exclusion between processes checking and setting the flag since it
cannot be interrupted during execution. The disadvantage is that this
technique is a form of busy waiting.
A better primitive method is semaphores (Dijkstra) which allows
processes, to be queued while they are waiting and restarted when the
shared resource (e.g. critical region) becomes available to them.
basic operations to the program are P (wait) and V (signal).
to be protected is bracketed as in
a.
wait(s)
[code
s ignal (s)
The
The code
Page
HARDWARE
f
F
W)
Two o(-
ihe
fbssVble
arvsv.ot.ri ar^
sV.;_.wy\
4
Page 5
2.
J-K Flip Flop
3_.
4-Bit Synchronous Binary Counter
«
Page 6
4
Architecture Concepts
a.
Multiprocessing means that there are several processors capable of
operating simultaneously on a system.
For example, swapping drums
and other I/O devices usually have their own processors which run
fairly independently of the cpu.
Some systems may even have
several cpu's.
b.
Memory interleaving means that successive addresses refer to words
which are physically located in different memory banks. Since
reading a word from memory is usually destructive, there is usually
a delay time for that bank during which local hardware restores the
contents of "the word. The effect of this delay can be minimized if
the next word fetched is from a different bank.
Another use for interleaving arises on memories having multiple
say for the cpu and swapping drum. Usually the I/O ports
ports
have higher priority.
If memory was not interleaved, a drum could
effectively lock the cpu (or some other device) out of a bank of
-
memory altogether during a rapid block transfer.
diminishes this
c.
Interleaving
"interference".
Wide buses permit the parallel transfer of more information at the
If each access to memory fetches two words at once, the
memory bandwidth is potentially doubled and a cpu may need only about half as
many bus cycles to carry out the instructions.
"Associative" memory means memory addressed by part of its content.
This is used to speed up the search for an entry in a table based on
same time.
d.
the parallel operation of the hardware which carries out the
association. It is often used to speed up the vi rtual -to-physical
address translation or in implementing a cache memory.
*
Page 7
THEORY OF COMPUTATION
1
.
Diagonal ization
a.
Assume that there Is an effective listing
JT
,
T
,
T
,
. . Y\ of
total
recursive functions and that T.(j) denotes the result of applying
'
th
function in the list.
Let us define a new total function F as follows:
F(x) = T (x) + 1 for all x.
Clearly F is total since each T. is total, and it is recursive since
argument jto
the i
T. (j) can be computed
— given
i and j
.
Furthermore F is not in the list because it differs at argument i for
each T. in the list,
i
Therefore the list is not complete.
Since this contradiction appears
with any complete listing of the total recursive functions, there can
be no such l isti ng.
b.
Every context sensitive language is recursive.
If every recursive
language were context sensitive we would have a contradiction ;of part (a)
C_,
be an effective listing of all the
Alternatively, letlc,. C
,
.Y\
fixed (countable) alphabet^.
could be an appropriate ordering of the sets of productions.)
context sensitive grammars, over a
(This
Let us define the set S as
| a' l
L(C.)
where a is some letter in _T
S=
a
1 is
not in
.
j
S is recursive because L(Cr) is recursive.
But S is not context sensitive since it is generated by no context
sensitive grammar (that is, S differs in at least one word from each
L(C.)).
Hence S is a recursive set
sensitive.
2.
Finite Automata
a.
{D>* {.}{o} *{x}
s.
*
over £ which is
not context
Page 8
d.
This finite machine of part (a) is essentially that part of a
scanner which recognizes the numeric constant and variable tokens
in a language. Final state 7 corresponds to an integer constant,
state
6 corresponds
to a
floating point constant (including the
which is only a decimal point!), and state 4 corresponds
to a variable name.
Of course, real compilers have somewhat more
complicated scanners which allow only finite length tokens, etc.
The "reversal" part might be used if the source is (for some
reason) to be read backwards. Also scanners use equivalent
strange one
determi n i st ie \mach \ nes
*
.
Page 9
3.
Context-free Language Quickies
1.
The language contains no strings of even length.
Proof:
be a shortest string of even length that can be parsed as
B
,
SBc
S
.
AaS
,
or
,
Then
or
cA has one of the forms
,
where the A
aSA
strings which are shorter than oC
,
2.
(In fact, all
for some k> 0 .)
S
B
string has length
3
->
SBc
The shortest
production
A
,
hence
c
,
Abß
,
,
hence have odd length.
is a contradiction.
4k+l
B
A
b ,
S denote parsable
a
,
e<,
Let
This
strings are of length
,
and c
occurs only in the
can appear in the middle if and
only if the string has a parse tree
S
Abß
SBc
a
The only four solutions are
by
*
aab
or
baa
.
aßcbß
where each
B
is replaced
Page 10
ALGORITHMS AND DATA STRUCTURES
1.
ALGOL
60
Problem
procedure RK cc
Beginning of program:
i°>
loop 7
RKlST(x,y,H,x2,y2) ; RKLST(x2,y2,H,xS,y3) .
check stepsize:
for k := 1 step 1 until n do
if comp (yl[k].y3[k], eta)
>
eps then
begin 6
c
go to check stepsize
end;
x
x := 3;
while not out
c
repeat ;
11
end RK
can't be eliminated without introducing extra
computation. But it isn't such a harmful one, and one would expect an
example at the end of a language definition to illustrate most of the
The remaining
go to
features of that language.
.
4
Page 11
2.
Tree Traversal
(F) A (B) C (G)
where (F) = FLM
(B) = (J) B (D) EX
(G) = G (N) 0
(J) = JRT
(D) = PDHIQ
(N) = NSU
So it comes toFLMAJRTB P D HIQEKCGNSUO
3..
Sparse Matrix
f i ,j 2)
a.
A sequential list containing entries (i, j, A
nonzero A [j ,jj
b.
An "orthogonal list" arrangement like that in Knuth Section 2.2.6
Same as b or (for example) a hash table of size M C750 which
c.
.
A^i.jl
finds
d.
given the key <i,j>.
IF the hash function has the
form (a? + bj) mod M, it will be possible to perform operations
(iii) and (iv) without too much delay.
10000 consecutive cells with A Ci,j7 in location
L + 100 i + j for some L
o
o
«
.
*
for
.
Page 12
Wume r i cal Analysis
1
.
Banded Linear Systems
1) ( a )
(b)
If partial pivoting is not used the band width
will
not be increased
Partial pivoting could increase the band width from
4m +
(c)
.
2m+l
to
significantly increasing the
1
(c)
Additionally.
number of arithmetic operations.
(d)
storage requirement and the
logic must be introduced to carry out the
searches for pivots which increases the complexity of the code and
adds to the machine time required.
2.
Arithmetic and error analysis
a.
Floating point and real arithmetic do not have the same arithmetic
For example, the floating point number system is not
properties.
associative with respect to
+
.
To see this consider two digit
In this case
base 10 arithmetic.
f 1 (10 + fl (-10 + 0.1 ))
-
0
but
fl (fl (10
b.
-
10) +0.1) = 0
Forward error analysis must be carried out in floating point arithmetic which as shown above in
associative.
2)a.
is, among other things, not
This is extremely involved if anything useful can be
obtained at all.
Inverse error analysis can be carried out in
real arithmetic and is therefore much easier.
produces
simpler
It also usually
/bounds which are easier to work with and
consequently of more utility.
3.
The rank problem
a.
If we use the
I
norm, II xII
._, I
. .max
I<i<n
x.
and the induced matrix norm we find that
where A
m
1
I
,x
~
),
= (x.,...,x
1
n"
cond (A)
>
n ~2
2
is the matrix on the left hand side of the equality.
Page
3. a (cont. )
If we let
x = (1,
bb
">>
A -1 b
=x
2
, 22_n) ' and
2 "n
■
b = (0,0,
13
. .. ,0,2 2_n ) '
we obtain
which yields
||A- I ||2
-
2 n
=
and
lIAII.
-
-1
IIA
Since
||*IL >||
AT\\\
m
- ||x„m
1
iiaAi_>
00
1
so
cond (A)
n
Thus for large
b
!!.
det (A)
- HAIIJIA" !^
1
this matrix,
n-2
> 2
A, is quite ill conditioned.
(
= 1 , Ais clearly nonsingular and the diagonal
elements are certainly not small. This example shows that
"nearly singular" for large
then very large.
n
A
is
since its condition number is
Methods based upon Guassian elimination to deter-
mine rank must rely upon the size of the diagonal elements, once
transformed to upper diagonal form, to determine rank.
r
rows have diagonals which are "essentially zero"
that the rank was
n-r
.
If the lower
we would conclude
Our example shows that there are nearly
singular matrices, or matrices which are only a small perturbation
from being singular which have large diagonal elements.
Therefore,
this approach will not be suitable for floating point computation.
«
Page 14
k.
Newton's Method
The iteration will clearly not converge and from the
picture it is reasonable to conclude that
=
X
n
X
n+2
=
=
x
=
"*'
n+2m
and
x
n+l
" X
n+3
X
n+2m+l
p(x)
Let
"
X
~
n+2m+l
'''
m so
for all
b.
~'
" X
n+2m
=Sn xn +
£
I= I
a.x
X
n+l
,
1
"Xnl>" * > °
- ,
p(r) =o,£r= r r
o<i<n-1
Then
£>(*")
n
= A a -?
n
* p( r )
and via Taylor expansion we obtain, to terms
p(r)
M
p(r) + Ar-p'(r)
.
Consequently,
n
-Aan mr
and to terms
or
«
,
= -A a *(r +
n
&&m) n » Ar«p'(r)
&(^r + Ar 'A an )
*r «A r 'P( r )
-A an
n
n-1
a r
w - V
_V_.
&r
A£
pTr) "
r
~-
Aa
-Ail
an
2
&(^r ),
A an
-a^
Page 15
.
.
4 ( cont )
c.
i- 1, 2, ...,10
For
P (x i )
Herman has observed that
P( X
i-1
m 1
}
2
p'Ui)
so
X
i+l
x.i
I
" X
i|
- x.,i-1
1
2
and the convergence is linear for these iterations.
This may have occurred because:
i) the Iterates are not close enough to the root, or
ii)
the
_
"*
x. are converging to a multiple root.
Page 16
ARTIFICIAL INTELLIGENCE
1.
Understanding Systems
(The following is an example of an acceptable
answer. )
"English Language Understanding"
A program which '^understands
English must have some way of
representing the meaning of sentences.
This meaning needs to be
incorporated into the program's mod e l of the world.
For example, In-
Winograd' s SHRDLU a model of the "blocks world" is kept and updated by
In Colby's PARRY, the variables of the program model the
the program.
emotions of the paranoid.
In Shank's work, a semantic net carries the
meaning in a data structure.
The importance of the model derives in part from the necessity of
the program to " r eason" about its world. For example SHRDLU knows
about the transitivity of "on top of" and also knows what it has done.
The MYCIN program knows "why" it made each step of a diagnosis. The
In
models in these cases include the prior activities of the program.
general A.I. developments, reasoning has been based on theorem proving
with new techniques involving the Resolution Principle. Older general
techniques include "means-ends analysis" of GPS. Today's more
successful programs use more domain specific techniques. SHRDLU
'reasons' about its model using the goal directed computations of
/"--PLANNER. In a sense, it knows "what follows from what" although
it does not solve theorems in general.
Like several of the new A.I.
languages, A-PLANNER does the book keeping of back tracking.
This tool
lets a program gracefully abandon a fruitless subgoal.
Sometimes there are several sources of knowledge for a program to
draw upon or there is knowledge to be used only in particular contexts.
The first case is called the "multiple expert" problem and is
exemplified by the HEARSAY program. The program subjugates the various
experts of an English-speaking chess playing program (sound expert,
syntax expert, chess expert, ...) to a MASTER which arbitrates between
them.
The second case is handled by a new software concept - the
pattern-directed (or pre-conditioned) interrupt known as a demon.
M
If
Page
17
the entire calculation is driven by patterns of the form
<pre-condi tion>-?<action> , then the system is termed a "production
system". These various concepts can be viewed either as fancy
developments in program organization, or as variations on the procedural
representation of knowledge.
In summary, English understanding is a rich enough area to require
and impact all the major areas of A.I. research, Including heuristic
search, knowledge representation, and new program organizations.
2
ERTS
The main A.I. problem for the design of the satellite's "understanding"
program is to determine specifically what objects are to be
specifically what description will be used for the
recognized:
object-finding; and specifically how the "finding" process will
exploit the description.
Thus, the satellite has to be able to
recognize specific objects such as rivers, river bends, deltas,
mountains, islands, etc., which it will report about; and effective
computational models of these (descriptions based on world knowledge)
constructed.
3
A. I
1.
.
Concepts
Heuristic Search. For any task regarded as nontrivial, the search
space is very large. Rules and procedures called heuristics are
applied to steer or limit the search. Often the heuristic rules
are greatly task specific.
Heuristic search is useful when no
formal search method exists (or is feasibly applied) or when informal
knowledge, i.e. heuristics, exist for pruning of and steering the
search space.
2.
Means-Ends Analysis
is a problem solving method applicable where a
goal state (G) is explicitly defined, an initial state (l) is
explicitly given, a set of operators is available to perform state
change, and a goal state can be achieved by incremental progress.
The method is to iteratively select an operator to transform
the current state into an intermediate state I' which is closer
to G.
*
The process stops when I
'
= G.
Page
3.
18
-
Hill Climbing
is a technique for exploring for the optima of a
function E of n variables. The technique generally cannot distinguish
between local and global optima
the "hiJls" of the surface in n+l
dimensions. The technique is useful as a performance improvement
method for the best performance setting of n variables where E
is the performance evaluation function. Unfortunately, "best" is
-
sometimes a molehill compared to nearby mountains.
4.
Demons
- a conditional
interrupt which operates autonomously in a
program.
It activates whenever the "condition" occurs in the
program. This conditional expression may be simple or a complicated
pattern matching expression. "Demons" are useful for representing
a process which can be conveniently conceptualized in terms of a few
independent pattern-invoked actions or subprocesses in addition to
the main computation.
5.
Production Systems
- are a
way of representing a program in terms
<s i tuation, action> pairs which are called productions. The
basic idea is that the action (subprocess or event) takes place when
the situation (pattern matching conditional expression) becomes true.
A more precise flow control (arbitration between competing actions)
is handled differently on different systems. This technique of
of
programming Is useful for representing processes which consist of
many independent pattern-invoked subprocesses.
The technique is
useful in a system where easy experimentation ?n terms of the
order of operations is required or where it is desired to add new
subprocesses (e.g. processing knowledge) without revising large portions
of the system.
3
A.!
1
Quickies
True - each application of an operator can be viewed as the reduction
of the problem to a simpler subproblem. The special case of state
space methods remains useful as a separate concept from problem
reduction because (l) some problems are more easily conceptualized
as "moves in a state space" than as "problem reduction steps";
(2) sometimes simpler algorithms apply.
(Prof. Knuth has remarked
that he has some especially efficient algorithms which apply in this
case.
._■
He cites a strong analogy between "state-space approach"
<—>"f i ni te
Page 19
state languages" and "problem reduction approach"**-*"context-free
languages".
-
- research
paper in progress)
the alpha-beta technique always achieves exactly the same
result as minimax.
2
False
3
Detailed knowledge of task domains has been found to be useful and
necessary for high levels of performance.
Attention has turned to
acquiring and utilizing such detailed knowledge for specific
problem domains. These particulars seem necessary to achieve
significant search reduction. In addition, general problem solving
methods have tended to force awkward, non-natural representations
of problems and domain-dependent knowledge.
*